manipulating a physics formula

**vcap420** · March 28th 2008, 06:37 AM

It's been a while since I've been in school (10yrs). I'm a little rusty at all my math. I wonder if someone could prove how the first formula is the same as the second.
Cheers!!!

**wingless** · March 28th 2008, 09:20 AM

$Vf^2 = Vi^2 +2a\Delta t$

Subtract $Vi^2$ from both sides

$Vf^2 - Vi^2 = 2a\Delta t$

Divide both sides by $2\Delta t$

$a = \frac{Vf^2 - Vi^2}{2\Delta t}$

I see that this is not the same as the second formula you posted. Yours was $a = \frac{Vf^2 - Vi^2}{2\boxed{a}\Delta t}$

I don't even have an idea what formula is this. It looks like the change of position formula of a constant acceleration, without time. Which is,
$\Delta x$ as the change of position,
$V_1$ and $V_2$ as the first and last velocities,
$a$ as the constant acceleration

$\Delta x = \frac{V_2^2-V_1^2}{2a}$

**topsquark** · March 28th 2008, 10:34 AM

Obviously, there was a typo in the second equation.

-Dan

**vcap420** · March 30th 2008, 06:01 AM

Your absolutely right there was a typo in the second equation. I was quite tired when I entered the equation. The equation involves distance rather than " t " time as well. My second lesson here is the importance of proof reading my equations. Thanks again.
Cheers!!!

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